Infinite Limit Proof

**Calc1stu** · October 26th 2008, 11:33 AM

I need to prove:

lim (x^4)-(2(x^2)) = +inf
x->+inf

Any suggestions?

**flyingsquirrel** · October 26th 2008, 11:52 AM

Hi,

Hint : $x^4-2x^2=x^4\left(1-\frac{2}{x^2}\right)$

**Calc1stu** · October 29th 2008, 12:46 PM

I don't know how to do it. The problem is done and over with, but I still would like to know how to get the answer for my own knowledge.

**Mathstud28** · October 29th 2008, 12:52 PM

Originally Posted by Calc1stu

I don't know how to do it. The problem is done and over with, but I still would like to know how to get the answer for my own knowledge.

$\lim_{x\to\infty}x^4\left(1-\frac{2}{x^2}\right)=\lim_{x\to\infty}x^4\cdot\lim _{x\to\infty}\left(1-\frac{2}{x^2}\right)=\infty\cdot{1}=\infty$

**Calc1stu** · October 30th 2008, 08:05 AM

It was supposed to be an epsilon delta proof.

**o_O** · October 30th 2008, 09:06 AM

You want to show that for every $M > 0$ , there exists $N > 0$ such that whenever $x > N$ , it necessarily follows that $f(x) > M \Leftrightarrow x^4 - 2x^2 > M$

Consider the latter expression as an equality:
$x^4 - 2x^2 = M$

$\Leftrightarrow x^4 - 2x^2 - M = 0$

$\Leftrightarrow x^2 = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-M)}}{2}$

$\Leftrightarrow x^2 = \frac{2 \pm \sqrt{4 + 4M}}{2}$

$\Leftrightarrow x^2 = 1 \pm \sqrt{1 + M}$

$\Leftrightarrow x^2 = 1 + \sqrt{1 + M}$ (since $x^2 \geq 0$ and $M > 0$ )

$\Leftrightarrow x = +\sqrt{1 + \sqrt{1 + M}}$ (we're considering positive x's)

So when $x \ {\color{red}=} \ \sqrt{1 + \sqrt{1 + M}}$ we have that $f(x)\ {\color{red} =}\ M$ . We want to show that for some $x \ {\color{magenta}>}\ N$ , we have that $f(x) \ {\color{magenta}>}\ M$

Now consider: $x \ {\color{magenta}> }\ \sqrt{1 + \sqrt{1 + M}} > 0$ . Can you finish?

**Calc1stu** · October 30th 2008, 11:19 AM

As far as epsilon and delta are concerned which is M and N. I don't know where to go from there. All we ever usually do or did was a basic function.

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